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Consider a finite wing at an angle of attack of \(6^{\circ}\). The normal and axial force coefficients are \(0.8\) and \(0.06\), respectively. Calculate the corresponding lift and drag coefficients. What comparison can you make between the lift and normal force coefficients?

Short Answer

Expert verified
Lift coefficient \( C_L \approx 0.7893 \), Drag coefficient \( C_D \approx 0.1433 \); lift is close to normal force coefficient.

Step by step solution

01

Understand the Given Quantities

We have a finite wing with an angle of attack \( \alpha = 6^{\circ} \). We are given the normal force coefficient \( C_n = 0.8 \) and the axial force coefficient \( C_a = 0.06 \). The goal is to find the lift coefficient \( C_L \) and the drag coefficient \( C_D \).
02

Recall Lift and Drag Coefficient Formulas

The lift coefficient \( C_L \) and drag coefficient \( C_D \) can be calculated as:\[ C_L = C_n \cos(\alpha) - C_a \sin(\alpha) \] \[ C_D = C_n \sin(\alpha) + C_a \cos(\alpha) \] where \( \alpha = 6^{\circ} \).
03

Convert Angle of Attack to Radians

Since trigonometric calculations require the angle in radians, convert \( 6^{\circ} \) into radians using the formula: \( \alpha_{radians} = \alpha_{degrees} \times \frac{\pi}{180} \). So, \( 6^{\circ} = 6 \times \frac{\pi}{180} = 0.1047 \) radians.
04

Calculate Trigonometric Functions

Find \( \cos(6^{\circ}) \) and \( \sin(6^{\circ}) \):- \( \cos(6^{\circ}) = \cos(0.1047) \approx 0.9945 \)- \( \sin(6^{\circ}) = \sin(0.1047) \approx 0.1045 \)
05

Calculate the Lift Coefficient

Substitute the values into the lift coefficient formula:\[ C_L = 0.8 \cdot 0.9945 - 0.06 \cdot 0.1045 \]\[ C_L \approx 0.7956 - 0.0063 = 0.7893 \]
06

Calculate the Drag Coefficient

Substitute the values into the drag coefficient formula:\[ C_D = 0.8 \cdot 0.1045 + 0.06 \cdot 0.9945 \]\[ C_D \approx 0.0836 + 0.0597 = 0.1433 \]
07

Compare Lift and Normal Force Coefficients

Since the lift coefficient \( C_L \approx 0.7893 \) is very close to the normal force coefficient \( C_n = 0.8 \), and both are significantly larger than the drag coefficient \( C_D \approx 0.1433 \), the lift force is the dominant component in comparison to the axial force, especially at small angles of attack like \( 6^{\circ} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lift Coefficient
The lift coefficient, denoted as \( C_L \), is a crucial parameter in aerodynamics that quantifies the lift produced by a wing or airfoil concerning the dynamic pressure and wing area. Lift is the force that opposes the weight of an aircraft and allows it to stay in the air.
  • The general formula to calculate the lift coefficient for a finite wing is \( C_L = C_n \cos(\alpha) - C_a \sin(\alpha) \).
  • Here, \( C_n \) is the normal force coefficient, \( C_a \) is the axial force coefficient, and \( \alpha \) represents the angle of attack.
In simpler terms, the lift coefficient accounts for the contribution of both the normal and axial forces to generate lift.
When a wing is tilted at an angle, the pressure difference between the top and bottom surfaces increases, creating more lift.
In our example, with a 6-degree angle of attack, the calculated lift coefficient is approximately 0.7893, which shows that most of the force on the wing contributes to lift.
Drag Coefficient
The drag coefficient, represented as \( C_D \), measures how much aerodynamic drag force is experienced by a shape moving through a fluid, like air. Drag is the resistance force that acts opposite to the direction of motion.
  • The drag coefficient for a wing can be calculated using: \( C_D = C_n \sin(\alpha) + C_a \cos(\alpha) \).
  • In this formula, \( C_n \) and \( C_a \) are the normal and axial force coefficients, respectively, while \( \alpha \) is the angle of attack.
The drag coefficient is influenced by the shape of the wing, its surface roughness, and the angle of attack.
In our specific case, with an angle of attack of 6 degrees, the drag coefficient is computed to be approximately 0.1433.
This value indicates that although drag is present, it remains much smaller compared to lift at lower angles of attack, showcasing efficient aerodynamic performance.
Finite Wing
A finite wing refers to real-world wings that have a definite span and are distinct from theoretical, infinitely long wings which don't exist in practice. Studying finite wings is essential for understanding how lift and drag are influenced by wing tips and their induced effects.
  • Finite wings experience wingtip vortices that create additional drag known as induced drag.
  • These vortices affect the airflow around the wing, influencing both lift and downwash.
Finite wings are crucial in realistic aerodynamic analysis and design because they consider practical factors like span, aspect ratio, and overall size.
The exercise demonstrates how lift and drag coefficients differ in finite wings compared to the assumptions made for infinite wings.
An angle of attack, such as 6 degrees, can further highlight how these coefficients are calculated, proving vital in aerodynamics when designing efficient wings and aircraft performance.

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Most popular questions from this chapter

The Cessna Cardinal, a single-engine light plane, has a wing with an area of \(16.2 \mathrm{~m}^{2}\) and an aspect ratio of \(7.31\). Assume that the span efficiency factor is 0.62. If the airplane is flying at standard sea-level conditions with a velocity of \(251 \mathrm{~km} / \mathrm{h}\), what is the induced drag when the total weight is \(9800 \mathrm{~N}\) ?

Consider a wing in a high-speed wind tunnel. At a point on the wing, the velocity is \(850 \mathrm{ft} / \mathrm{s}\). If the test-section flow is at a velocity of \(780 \mathrm{ft} / \mathrm{s}\), with a pressure and temperature of \(1 \mathrm{~atm}\) and \(505^{\circ} \mathrm{R}\), respectively, calculate the pressure coefficient at the point.

Consider a rectangular wing with a NACA 0009 airfoil section spanning the test section of a wind tunnel. The test-section airflow conditions are standard sea level with a velocity of \(120 \mathrm{mi} / \mathrm{h}\). The wing is at an angle of attack of \(4^{\circ}\), and the wind tunnel force balance measures a lift of \(29.5 \mathrm{lb}\). What is the area of the wing?

Consider a flat plate oriented at a \(90^{\circ}\) angle of attack in a low- speed incompressible flow. Assume that the pressure exerted over the front of the plate (facing into the flow) is a constant value over the front surface, equal to the stagnation pressure. Assume that the pressure exerted over the back of the plate is also a constant value, but equal to the free-stream static pressure. (In reality, these assumptions are only approximations to the real flow over the plate. The pressure over the front face is neither exactly constant nor exactly equal to the stagnation pressure, and the pressure over the back of the plate is neither constant nor exactly equal to the free-stream pressure. The preceding approximate model of the flow, however, is useful for our purpose here.) Note that the drag is essentially all pressure drag; due to the \(90^{\circ}\) orientation of the plate, skin friction drag is not a factor. For this model of the flow, prove that the drag coefficient for the flat plate is \(C_{D}=1\).

In some aerodynamic literature, the drag of an airplane is couched in terms of the "drag area" instead of the drag coefficient. By definition, the drag area, \(f\), is the area of a flat plate at \(90^{\circ}\) to the flow that has a drag force equal to the drag of the airplane. As part of this definition, the drag coefficient of the plate is assumed to be equal to 1 , as shown in Prob. \(5.31\). If \(C_{D}\) is the drag coefficient of the airplane based on wing planform area \(S\), prove that \(f=C_{D} S\).

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